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C(s) + O2(g) -> CO2(g) ΔH = -394 kJ/mol
H2(g) + 1/2 O2(g) -> H2O(l) ΔH = -286 kJ/mol
4C(s) + 5H2(g) -> C4H10 (g) ΔH = -126 kJ/mol

The standard enthalpy of combustion of butane, in kJ/mol, is:

A -2880
B -2590
C -806
D -554

Could someone show me how to do this question? Thank you!

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5 answers
  1. eqn 3 reversed + 4*eqn 1 + 5*eqn 2

    Post your work if you get stuck.

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  2. So the answer is A, right? Thanks for the help!

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  3. I didn't do the math.

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  4. Yes, A is correct.

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  5. Hey! That's pretty goooood!

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